If the mean of x is 0, is the mean of x-squared also 0?

[Apashanka]

if $\bar x =\frac{\int x dv}{\int dv}=0$
Is necessarily $\bar {x^2}=0$??

 

[Doc Al]

Note that while $x$ may be negative, $x^2 \geq 0$.

 

[Dale]

Let $x=\{-1,1\}$.

Then $\bar x=0$ but $\bar{x^2}=1$

 

[Apashanka]

Is the reverse true, e.g if $\bar {x^2}=0$ then $\bar x=0$??, given $x>0$ and
$\bar x=\frac{\int x dv}{\int dv}$

 

[Nugatory]

Is the reverse true, e.g if $\bar {x^2}=0$ then $\bar x=0$??, given $x>0$ and
$\bar x=\frac{\int x dv}{\int dv}$

What do you think and why?

 

[Apashanka]

What do you think and why?

Because I came across a situation where it is given that as $\bar {A^2}$=0 ,so $\bar {\vec A}$=0,that why I am asking

 

[sophiecentaur]

Because I came across a situation where it is given that as $\bar {A^2}$=0 ,so $\bar {\vec A}$=0,that why I am asking

"Situation"? That could only be true if all values of A are zero or complex, I think.

 

[boneh3ad]

Just to be clear, do you mean $\overline{A^2_i}$? The rendering for \bar{} is terrible in $\LaTeX$. Consider using \overline{} instead.

 

[Apashanka]

For $\vec A$ if $\overline {A^2}=0$ implies that $\overline{A_i^2}=0$ using this $\frac{A_i \int_V A_i dv}{\int_V dv}-\frac{\int_V (\frac{dA_i}{dv}\int_V(A_i dv))dv}{\int_V dv}(=\bar A_i \int_V dA_i)=0$ can this be used to proof that $\bar A_i=0??$
(Average is taken over a volume V and $\bar A_i=\frac{\int_V A_i dv}{\int_V dv}$)

 

[Ray Vickson]

For $\vec A$ if $\overline {A^2}=0$ implies that $\overline{A_i^2}=0$ using this $\frac{A_i \int_V A_i dv}{\int_V dv}-\frac{\int_V (\frac{dA_i}{dv}\int_V(A_i dv))dv}{\int_V dv}(=\bar A_i \int_V dA_i)=0$ can this be used to proof that $\bar A_i=0??$
(Average is taken over a volume V and $\bar A_i=\frac{\int_V A_i dv}{\int_V dv}$)

The equation $\overline{A_i^2}=0$ means that
$$\frac{\int_V A_i^2 \, dv}{\int_V \, dv} = 0.$$ So, if the $A_i$ are real numbers, this implies $A_i = 0$ almost everywhere, and so $\bar{A_i} = 0$ also.

 

[WWGD]

For $\vec A$ if $\overline {A^2}=0$ implies that $\overline{A_i^2}=0$ using this $\frac{A_i \int_V A_i dv}{\int_V dv}-\frac{\int_V (\frac{dA_i}{dv}\int_V(A_i dv))dv}{\int_V dv}(=\bar A_i \int_V dA_i)=0$ can this be used to proof that $\bar A_i=0??$
(Average is taken over a volume V and $\bar A_i=\frac{\int_V A_i dv}{\int_V dv}$)

What kind of Mathematical object is your $A_i$?

 

[Apashanka]

What kind of Mathematical object is your $A_i$?

$A_i$ are components of a vector

 

[WWGD]

$A_i$ are components of a vector

But how do you define the integral in this case? Is it given as a Real- or Otehrwise- valued function?

 

[Apashanka]

But how do you define the integral in this case? Is it given as a Real- or Otehrwise- valued function?

It's real and of course positive which are functions of $f(x,y,z,t)$

 

[WWGD]

It's real and of course positive which are functions of $f(x,y,z,t)$

EDIT: Well, if the (Riemann) integral of a positive Real-valued function is 0, then the function must be identically 0. Otherwise, if $f \geq \epsilon \geq 0$ then $\int_{[a,b]} f \geq \epsilon(b-a) >0$

EDIT2: At most f can be non-zero at a set of measure 0.

 

[Apashanka]

EDIT: Well, if the (Riemann) integral of a positive Real-valued function is 0, then the function must be identically 0. Otherwise, if $f \geq \epsilon \geq 0$ then $\int_{[a,b]} f \geq \epsilon(b-a) >0$

EDIT2: At most f can be non-zero at a set of measure 0.

What I am just thinking( say in cartesian coordinates) is $\int_v f(x,y,z)dv=\int_v f(x,y,z)dxdydz=\int_x f_x(x)dx\int_y f_y(y)\int_z f_z(z)dz$this expression is zero if each of the area under the curve on the $x,y,z$ axis is 0 ,or if it is having a point of zero crossing.
For the function to be positive valued the first condition above is to be satisfied,isn't it??

 

[haushofer]

if $\bar x =\frac{\int x dv}{\int dv}=0$
Is necessarily $\bar {x^2}=0$??

Besides Dale's example: Wouldn't that imply that a mean of 0 would imply a variance of 0? That would be really odd, I'd say.

 

[Apashanka]

Besides Dale's example: Wouldn't that imply that a mean of 0 would imply a variance of 0? That would be really odd, I'd say.

Yes it's variance is then coming 0

 

[WWGD]

What I am just thinking( say in cartesian coordinates) is $\int_v f(x,y,z)dv=\int_v f(x,y,z)dxdydz=\int_x f_x(x)dx\int_y f_y(y)\int_z f_z(z)dz$this expression is zero if each of the area under the curve on the $x,y,z$ axis is 0 ,or if it is having a point of zero crossing.
For the function to be positive valued the first condition above is to be satisfied,isn't it??

I don't think you can always find a primitive. But if you did, since f (x,y,z) is Real valued, it would evaluate as
F(x_1,y_1,z_1)-F(x_0,y_0,z_0)

Yes it's variance is then coming 0

Variance 0 implies all values are equal.

 

[WWGD]

What I am just thinking( say in cartesian coordinates) is $\int_v f(x,y,z)dv=\int_v f(x,y,z)dxdydz=\int_x f_x(x)dx\int_y f_y(y)\int_z f_z(z)dz$this expression is zero if each of the area under the curve on the $x,y,z$ axis is 0 ,or if it is having a point of zero crossing.
For the function to be positive valued the first condition above is to be satisfied,isn't it??

I don't think you can always find a primitive. But if you did, since f (x,y,z) is Real valued, it would evaluate as
F(x_1,y_1,z_1)-F(x_0,y_0,z_0)

I don't think you can always find a primitive. But if you did, since f (x,y,z) is Real valued, it would evaluate as
F(x_1,y_1,z_1)-F(x_0,y_0,z_0)
Variance 0 implies all values are equal.

Expanding on this, yes, if your volume is zero, you can see it as having "degenerate" volume, i.e., a figure that is not strictly 3D ( under the same assumption that f(x,y,z) is non-negative), and that allows for degeneracy along different dimensions.

 

[Nik_2213]

Without wandering into calculus, I'm reminded of the statistics 'standard deviation' algorithm where you sum a list items, and also sum the squares of those list items...

 

[Master1022]

if $\bar x =\frac{\int x dv}{\int dv}=0$
Is necessarily $\bar {x^2}=0$??

Not necessarily, think about sin(x) and cos(x)

 

[Michael Hardy]

The variance of a real-valued random variable is the mean of the square of something whose mean is $0$. If the proposed proposition were true, then every variance would be $0,$ and that is nonsense. The example already pointed out, where the two possible values of $x$ are $\pm1,$ each with probability $1/2,$ may be the simplest example.

However, it is true that if the average value of the square of $x$ is $0$ and $x$ is real, then the average value of $x$ is $0,$ and in fact the probability that $x\ne0$ in that case is $0$.

 

[WWGD]

Ultimately there is the counter of the standard normal. $E[x]=0; \sigma=E[(x-0)^2]=E[x^2]=1\neq (E[x])^2=0^2=0$.

 

[epenguin]

For the initial question isn't this one of those cases where tn suffices to think of a simple example? For example for the mean of two quantities to be zero they have to be equal and opposite sign. The squares are also equal but they are not opposite. Proved except for unineresting qualification where everything is zero.

 

[WWGD]

For the initial question isn't this one of those cases where tn suffices to think of a simple example? For example for the mean of two quantities to be zero they have to be equal and opposite sign. The squares are also equal but they are not opposite. Proved except for unineresting qualification where everything is zero.

Yes, that is what we did. Take X~N(0,1). Then $E[X]=0, E[X^2]=E[(X-0)^2]=1 \neq E[X]^2$